Equivalent Expressions: Write, Test and Find Mistakes | Grade 6 Math | LevelKite

Three friends each buy one ticket and one snack. Let $x$ be the price of one ticket in dollars. The letter $x$ is a variable: it stands for a number that can change. Each snack costs $2$ dollars.

You can write the total cost in dollars as an expression, a math phrase built from numbers, variables and operations. Here are two ways. Count person by person: each friend pays $x + 2$ dollars, so the total is $3(x + 2)$ . Count item by item: three tickets cost $3x$ dollars and three snacks cost $6$ dollars, so the total is $3x + 6$ .

Both expressions count the same money, so they give the same total for every ticket price $x$ . Expressions like these are called equivalent.

In this lesson you will write equivalent expressions, test a pair of expressions with numbers, and find the mistake when a rewritten expression is not equivalent to the original.

Key concept

Equivalent means equal for every value

Two expressions are equivalent when they have the same value for every value of the variable in the stated domain. The domain is the set of values the variable may take, such as the possible ticket prices.

A property of operations is true for every number, so an expression rewritten by following a property is equivalent to the original.
An equality is a statement with an equal sign. It is preserved, meaning it stays true, when you apply the same valid operation to both sides. Divide only by a nonzero quantity.
Substitution checks one value at a time. A value in the domain where the two expressions differ is a counterexample. It proves the expressions are not equivalent. A match at one value does not prove they are equivalent.

Start with writing equivalent expressions. A property is a fact about addition or multiplication that is true for every number. That is where its power comes from. Rewrite an expression by following a property, and the new expression has the same value as the old one for every value of the variable in the domain. You do not have to test any numbers. In this lesson, the new expression is called a rewrite.

You met one property already. Counting the ticket money person by person or item by item is the distributive property at work. The table lists it with the other properties you will use. In each row, $x$ stands for any number, so the domain is all numbers.

These properties are about addition and multiplication. They do not let you change the order or the grouping in a subtraction or a division.

Properties of addition and multiplication, each true for every number

Property	What it says	Example
Commutative property of addition	You can change the order of the numbers you add. The sum stays the same.	$x + 9 = 9 + x$
Commutative property of multiplication	You can change the order of the numbers you multiply. The product stays the same.	$x \cdot 4 = 4 \cdot x$
Associative property of addition	You can change how the numbers you add are grouped. The sum stays the same.	$(x + 2) + 5 = x + (2 + 5)$
Associative property of multiplication	You can change how the numbers you multiply are grouped. The product stays the same.	$3 \cdot (2 \cdot x) = (3 \cdot 2) \cdot x$
Distributive property	Multiplying a sum by a number equals multiplying each part of the sum by that number, then adding the products.	$4(x + 5) = 4 \cdot x + 4 \cdot 5$

Two ideas help you use these properties inside longer expressions. Here is the first. Let $x$ stand for any number.

You can rewrite one part. An equality has two sides, one on each side of the equal sign. The distributive property tells you that $4(x + 5) = 4x + 20$ for every number $x$ . Now add $3$ to both sides. You applied the same valid operation to both sides, so the equality is preserved: $4(x + 5) + 3 = 4x + 20 + 3$ . This is why you may rewrite one part of an expression and copy the rest unchanged.

Which operations are valid? You may add the same quantity, such as $3$ or $4x$ , to both sides. You may subtract the same quantity from both sides, or multiply both sides by the same quantity. You may divide both sides by the same quantity only when that quantity is not zero.

Here is the second idea.

You can run the distributive property backward. You have done this before: it is factoring. In $12x + 18$ , the terms are the parts being added, $12x$ and $18$ . Both terms have the common factor $6$ , a number that divides each term evenly. Divide each term by $6$ to find what stays inside the parentheses: $12x \div 6 = 2x$ and $18 \div 6 = 3$ . So $12x + 18 = 6(2x + 3)$ for every number $x$ .

Why is this valid? Divide only by a nonzero quantity, and here the divisor is $6$ , which is not zero. Dividing by $6$ undoes multiplying by $6$ . So when you distribute the $6$ again, you get back the original: $6 \cdot 2x + 6 \cdot 3 = 12x + 18$ .

Worked example

Write equivalent expressions with properties

Let $x$ stand for any number. Write two expressions that are equivalent to $3(2x + 5) + 4x$ . Give one with no parentheses and as few terms as possible, and one that is factored. Name the property used in each step.

Use the distributive property: multiply each term inside the parentheses by 3. The 4x is added on both sides of the equal sign, so the equality is preserved.
$3(2x + 5) + 4x = 3 \cdot 2x + 3 \cdot 5 + 4x$
Do each multiplication. The associative property of multiplication lets you regroup 3 times 2x as (3 times 2) times x, which is 6x. Also, 3 times 5 is 15.
$3 \cdot 2x + 3 \cdot 5 + 4x = 6x + 15 + 4x$
The commutative and associative properties of addition let you add in any order. Swap 15 and 4x so the like terms, the terms with the same variable part, are next to each other.
$6x + 15 + 4x = 6x + 4x + 15$
Combine the like terms. This is the distributive property run backward: 6 groups of x plus 4 groups of x make 6 + 4 groups of x, which is 10 groups of x.
$6x + 4x + 15 = (6 + 4)x + 15 = 10x + 15$
Factor to get a second form. This is the distributive property run backward again. Both terms have the common factor 5. Divide each term by 5, which is valid because 5 is not zero: 10x divided by 5 is 2x, and 15 divided by 5 is 3.
$10x + 15 = 5(2x + 3)$

Answer

For every number $x$ , both $10x + 15$ and $5(2x + 3)$ are equivalent to $3(2x + 5) + 4x$ .

Properties build equivalent expressions. Substitution lets you test a pair of expressions. To substitute, replace the variable with a number from the domain. Use the same number in both expressions, then work out each value.

The two possible results do not tell you the same amount.

The values are different. You have found a counterexample: a value in the domain where the two expressions disagree. Equivalent expressions must agree for every value in the domain, so one counterexample proves the expressions are not equivalent.
The values match. You know the expressions agree at that one value. The other values in the domain are still unchecked, so a match at one substitution does not prove equivalence.

Testing more values can catch more mistakes. But to say yes for every value, you need a reason that covers every value. That reason comes from the properties.

Worked example

Test two candidates with substitution

Let $x$ stand for any number. Which expression is equivalent to $2(3x + 4)$ : expression A, $6x + 4$ , or expression B, $6x + 8$ ? Test both with substitution, then give a reason for your answer.

Choose a number from the domain, such as 2. Substitute it for x in the original expression. Work inside the parentheses first, then multiply.
$2(3 \cdot 2 + 4) = 2(6 + 4) = 2 \cdot 10 = 20$
Substitute the same number, 2, for x in expression A. The value is 16, but the original gave 20. That is a counterexample, so A is not equivalent to the original. You do not need to test A again.
$6 \cdot 2 + 4 = 12 + 4 = 16$
Substitute 2 for x in expression B. The value is 20, the same as the original. A match at one value does not prove equivalence, so you still need a reason that covers every number.
$6 \cdot 2 + 8 = 12 + 8 = 20$
Use the distributive property on the original: multiply each term inside the parentheses by 2. The result is expression B. A property is true for every number, so B is equivalent to the original.
$2(3x + 4) = 2 \cdot 3x + 2 \cdot 4 = 6x + 8$

Answer

Expression B, $6x + 8$ , is equivalent to $2(3x + 4)$ for every number $x$ , by the distributive property. Expression A, $6x + 4$ , is not equivalent: at $x = 2$ it gives $16$ , but the original gives $20$ .

Testing the rewrite at four values of the variable

Value of the variable	Original expression	Rewrite	Same value?
$x = 0$	$3 \cdot 0 + 4 = 4$	$7 \cdot 0 = 0$	No
$x = 1$	$3 \cdot 1 + 4 = 7$	$7 \cdot 1 = 7$	Yes
$x = 2$	$3 \cdot 2 + 4 = 10$	$7 \cdot 2 = 14$	No
$x = 3$	$3 \cdot 3 + 4 = 13$	$7 \cdot 3 = 21$	No

A counterexample tells you that a rewrite is wrong. The next job is to find where it went wrong. Look for the step that no property supports.

Take the rewrite of $3x + 4$ as $7x$ . The student added the $3$ and the $4$ . But $3x$ means $3$ groups of $x$ , and $4$ is a plain number. They are not like terms, because they do not have the same variable part. So the distributive property cannot join them. Compare that with $3x + 4x$ . Both terms have the factor $x$ , so $3x + 4x = (3 + 4)x = 7x$ for every number $x$ .

To diagnose any rewrite, use three moves.

Go line by line. Find the first line that does not follow from the line before it by a property.
Confirm with a counterexample: a value in the domain where the original and the rewrite have different values.
Fix that step by using the property correctly.

Worked example

Find and fix an invalid rewrite

Let $x$ stand for any number. Jordan rewrites $5(x + 3) + 2x$ in two lines. Line 1: $5x + 3 + 2x$ . Line 2: $7x + 3$ . Find Jordan's mistake, show that $7x + 3$ is not equivalent to the original expression, and write a correct equivalent expression.

Compare line 1 with the original. The distributive property says to multiply each term inside the parentheses by 5. Jordan multiplied the x by 5 but not the 3, so line 1 holds a mistake. The correct product is shown here.
$5(x + 3) = 5 \cdot x + 5 \cdot 3 = 5x + 15$
Compare line 2 with line 1. Jordan moved the like terms 5x and 2x together and combined them into 7x. Both moves follow the properties, so line 1 holds the only mistake.
$5x + 3 + 2x = 5x + 2x + 3 = 7x + 3$
Confirm with a counterexample. Choose a number from the domain, such as 2, and substitute it for x in the original expression.
$5(2 + 3) + 2 \cdot 2 = 5 \cdot 5 + 4 = 25 + 4 = 29$
Substitute the same number, 2, for x in Jordan's result. The value is 17, but the original gave 29. The values are different, so the two expressions are not equivalent.
$7 \cdot 2 + 3 = 14 + 3 = 17$
Fix line 1 with the distributive property: multiply both terms inside the parentheses by 5. The 2x is added on both sides, so the equality is preserved. Then move the like terms together and combine them.
$5(x + 3) + 2x = 5x + 15 + 2x = 5x + 2x + 15 = 7x + 15$
Check the fixed expression by substituting 2 for x. It gives 29, the same as the original. This match is not the proof. The properties used in step 5 are what make the expressions equivalent for every number x.
$7 \cdot 2 + 15 = 14 + 15 = 29$

Answer

Jordan multiplied only the $x$ by $5$ and not the $3$ . The result $7x + 3$ is not equivalent to $5(x + 3) + 2x$ , because at $x = 2$ the values are $17$ and $29$ . A correct equivalent expression is $7x + 15$ , for every number $x$ .

Here is the idea to keep. Equivalent expressions have the same value for every value of the variable in the stated domain.

To write an equivalent expression, or to explain why two expressions are equivalent, follow the properties. An equality is preserved when you apply the same valid operation to both sides, and you divide only by a nonzero quantity.
To test a pair, substitute the same value from the domain into both expressions. One counterexample proves they are not equivalent. A match at one value does not prove they are equivalent.
To find a mistake, look for the step that no property supports, confirm it with a counterexample, and fix it by using the property correctly.